The Stacks project

Instead of 26.4.6, I think one could better invoke 29.2.2.

I think the following folklore (that #5172 hints at) could be an interesting remark to add to this section: Given a locally ringed space $X$, there is a one-to-one correspondence between ideal sheaves over $X$ locally generated by sections and closed immersions of l.r.s. into $X$, modulo isomorphism over $X$. The map to the right sends $\mathcal{I}\subset\mathcal{O}_X$ to the closed subspace of $X$ associated to this sheaf of ideals (Definition 26.4.4), whereas the map to the left sends a closed immersion $f:Z\to X$ to $\operatorname{Ker}(\mathcal{O}_X\to f_*\mathcal{O}_Z)$. The composite "first to the right, then to the left" is the identity by 26.4.3, whereas the composite "first to the left, then to the right" is the identity by 26.4.5. Moreover, this one-to-one correspondence promotes to a contravariant isomorphism of posets when the ideal sheaves are ordered by inclusion and the order in the closed subspaces of $X$ is given by the existence of a map $Z\to Z'$ over $X$ (by monicity of closed immersions, there is at most one such map). On the one hand, if we have ideal sheaves $\mathcal{I}'\subset\mathcal{I}\subset\mathcal{O}_X$, then we get a map $Z\to Z'$ over $X$ by 26.4.6, where $Z$ and $Z'$ are the closed subspaces associated to $\mathcal{I}$ and $\mathcal{I}'$, respectively. On the other hand, if we have closed subspaces $f:Z\to X$ and $f':Z'\to X$ and a map $g:Z\to Z'$ over $X$, then $\mathcal{I}$ equals the kernel of $\mathcal{O}_X\to f_*\mathcal{O}_Z$, i.e., of $\mathcal{O}_X\to f'_*\mathcal{O}_{Z'}\to f_*'g_*\mathcal{O}_X$. But this contains $\mathcal{I}'$.

Also, at the beginning of the section, maybe one could inform the reader that the condition (3) of Definition 26.4.1 will not be ever used throughout the section? Everything stated in this section is still true if we were to drop (3) from this definition, and the proofs don't change (condition (3) is never invoked). I am aware of the discussion in Modules, Section 17.13, and I don't disagree with it. I just think it would be informative to have some note in this section regarding this issue.

Here's the proof of cartesianity of the square from the statement: we need to use

Exercise. Let $f:X\to Y$ be a morphism of ringed spaces and let $\varphi$ be a morphism of $\mathcal{O}_Y$-modules. Then $f^*\operatorname{Im}\varphi=\operatorname{Im}f^*\varphi$. (Hint: verify it on stalks and use that if $R$ is a commutative ring, then $L\otimes\operatorname{Im}\psi=\operatorname{Im}(L\otimes\psi)$, for $L$ an $R$-module and $\psi$ a morphism of $R$-modules.)

Suppose we have morphisms $g:W\to X$ and $h:W\to Z$ fitting into a commutative square which terminates in $Y$. If a factorization of this square through $Z'$ were to exist, then it would be unique for $i$ is monic (closed immersions of l.r.s. are monomorphisms). We see existence: We construct a map $W\to Z'$ factoring $g$ through $i'$ with 26.4.6: we must show that the map $$g^*\operatorname{Im}(f^*\mathcal{I}\to f^*\mathcal{O}_Y)\to g^*\mathcal{O}_X$$ vanishes. But we have $$\begin{align*} g^*\operatorname{Im}(f^*\mathcal{I}\to f^*\mathcal{O}_Y) &=\operatorname{Im}(g^*f^*\mathcal{I}\to g^*f^*\mathcal{O}_Y), &\text{by the exercise,}\\ &=\operatorname{Im}(h^*i^*\mathcal{I}\to h^*i^*\mathcal{O}_Y)\\ &=0, \end{align*}$$ by 26.4.6. On the other hand, the induced map $W\to Z'$ can be seen to also factor $h$ through $f'$ using the monicity of $i$.

Slogan: closed immersions of locally ringed spaces are stable under base change.

I think my comment was wrong. There is a filtration after base change, and you can use it to prove the result. However, that is not the shortest proof, and I do not know if the filtration comes from a sequence of sections. I need to double-check what is in "N'eron models" -- they have effectively evicted us from our offices while they repair the HVAC system.

Thanks Jason!

Since we've defined the normalization only for a quasi-compact and quasi-separated morphism, shouldn't one remark at some point during the proof that this is indeed the case for $g$? (It follows from 26.21.13 and 26.21.14.)

Also, instead of "with a unique morphism $h:X'\to Z$. We omit the verification that the diagram commutes," one could write "with a unique morphism $h:X'\to Z$ over $X$. On the other hand, by https://stacks.math.columbia.edu/tag/01LQ#comment-8448 , we have $h\circ f'=g$."

I propose to add the following remark to the webpage: Let $S$ be a scheme and let $\varphi:\mathcal{A}\to\mathcal{B}$ be a morphism of quasi-coherent $\mathcal{O}_S$-algebras. As it was explained in #8438 (ii), we obtain a morphism $\underline{\text{Spec}}_S\mathcal{B}\to \underline{\text{Spec}}_S\mathcal{A}$ over $S$. This gives a transformation of functors $\text{Hom}_S(-,\underline{\text{Spec}}_S\mathcal{B})\to\text{Hom}_S(-,\underline{\text{Spec}}_S\mathcal{A})$ that induce a transformation of functors $G'\to G$, where $G$ is our old friend and $G'$ is the functor taking a scheme $f:T\to S$ over $S$ to $\text{Hom}_{\mathcal{O}_S\text{-Alg}}(\mathcal{B},f_*\mathcal{O}_T)$. We claim that at $f:T\to S$ the transformation $G'\to G$ is given by precomposition by $\varphi$. We begin with a morphism $\mathcal{B}\to f_*\mathcal{O}_X$ of $\mathcal{O}_S$-algebras that gives a morphism $g:T\to\underline{\text{Spec}}_S\mathcal{B}$ over $S$. In turn, we get a composite $T\to\underline{\text{Spec}}_S\mathcal{B}\to\underline{\text{Spec}}_S\mathcal{A}$. By #8438 (i), this gives a morphism of $\mathcal{O}_S$-algebras $\mathcal{A}\to f_*\mathcal{O}_T$ that on sections over an open affine $U\subset S$ reads as $\mathcal{A}(U)\xrightarrow{\varphi}\mathcal{B}(U)\to\mathcal{O}_X(f^{-1}U)$, i.e., it equals the composite $\mathcal{A}\to\mathcal{B}\to f_*\mathcal{O}_X$, which is what we wanted to show.

By the discussion in "N'eron models" around finite flat morphisms, after a faithfully flat base change there is a filtration of f_*O_X whose associated graded pieces are f-pushforwards of pushforwards of invertible O_S-modules along sections of f (i.e., multisections are "scheme-theoretic unions" of sections after faithfully flat base change). The composition of those sections with sigma give a corresponding filtration of the structure sheaf of the relative Proj (by ideal sheaves) compatible with the filtration on the pushforward of O_X. You can use this to prove that the morphism from the structure sheaf of the relative Proj to the pushforward of O_X is surjective.

I think instead of "we have $a_i \in R_f$ and $x^d + \sum_{i = 1, \ldots, d} \varphi(a_i) x^{d - i} = 0$ in $S_f$" it would be more informative to write "there are $b_i\in R_f$ such that $x^d + \sum_{i = 1, \ldots, d} \varphi(b_i) x^{d - i} = 0$ in $S_f$." At least in the calculation I derived I end up with different $a_i$'s (this doesn't affect the rest of the proof).

I agree with MAO Zhouhang that it's useful to mention here that $D_{I^\infty\text{-}\mathrm{torsion}}(A)$ is the category of $A$ complexes with $I^\infty$-torsion cohomology.

Argh, the reference to Lemma 26.21.11 is still wrong because the Proj is over $S$ and not over $X$. The best thing would be for people to work out for themselves why $\sigma$ is a closed immersion (by doing a local computation). But we can also see it using theory. Write $X$ as $\text{Proj}_S(\mathcal{A})$ where $\mathcal{A} = (f_*\mathcal{O}_X)[T]$. (Namely, we have $\text{Proj}(S[T]) = \text{Spec}(S)$ and there is a relative version of this.) Then with $\mathcal{B} = \text{Sym}^*_{\mathcal{O}_S}(f_*\mathcal{O}_X)$ there is a map $\psi : \mathcal{B} \to \mathcal{A}$ of graded $\mathcal{O}_S$-algebras. (Namely, given a ring map $R \to S$ there is a graded $R$-algebra map $\text{Sym}_R^*(S) \to S[T]$, $s \mapsto sT$ and there is a relative version of this.) The map $\psi$ is surjective in all sufficiently high degrees. (Namely, in degrees $\geq 1$.) Whence the corresponding morphism $r_\psi : X \to \mathbf{P}(f_*\mathcal{O}_X)$ is everywhere defined and a closed immersion, see Lemma 27.18.3.

Okay, sorry, the conditions $\rho_U^U=\text{id}_{X_U}$ and $\theta^U_U=\text{id}_{\mathcal{F}_U}$ already follow from the current hypotheses (maybe one could mention this?) .

In 27.2.2.1, I think the middle and last arrows should be labelled as $(i_V^{-1})^*(\theta_U|_{f^{-1}(V)})$ and $(i_V^{-1})^*\theta_V^{-1}$, respectively.

Instead of saying "by covering the intersection $U_1\cap U_2=\bigcup V_j$ by elements $V_j$ of $\mathcal{B}$ and taking" I think it would be better to write "by considering any $V\in\mathcal{B}$ such that $V\subset U_1\cap U_2$ and defining", since actually using some covering is not enough to later construct $\varphi_{12}$.

To give just the minimum amount of hints to avoid the first use of "we omit," one could:

1. After "condition (d) says exactly that this is compatible in case we have a triple of elements $W\subset V\subset U$ of $\mathcal{B}$," add "i.e., $(\theta')^U_W=(\theta')^V_W\circ(\theta')^U_V|_{f^{-1}(W)}$."
2. If we used the notation $\varphi^{V_j}_{12}$ for what the proof currently denotes $\varphi_{12}|_{V_j}$, then instead of "we omit the verification that these maps do indeed glue to a $\varphi_{12}$" one could write "since $\varphi^V_{12}|_W=\varphi^W_{12}$ for $V,W\in\mathcal{B}$ with $W\subset V\subset U_1\cap U_2$, by Sheaves, 6.33.1, we get a $\varphi_{12}$."

On the other hand, I think that for 27.2.2.1 to follow, in the lemma hypotheses we would need to add "suppose $\rho_U^U=\text{id}_{X_U}$ and $\theta^U_U=\text{id}_{\mathcal{F}_U}$ for $U\in\mathcal{B}$." The proof of 27.2.2.1 may be added after "we get our $\mathcal{F}$ on $X$" by adding "along with isomorphisms $\theta_U:\mathcal{F}'_U\to\mathcal{F}|_{f^{-1}(U)}$, $U\in\mathcal{B}$, such that $\theta_U|_{f^{-1}(U\cap V)}=\theta_V|_{f^{-1}(U\cap V)}\circ \varphi_{U,V}$ for $U,V\in\mathcal{B}$. In particular, for $U,V\in\mathcal{B}$ with $V\subset U$, we have $\theta_V^{-1}\circ\theta_U|_{f^{-1}(V)}=\varphi_{U,V}=(\theta')^U_V=i^*_V\theta^U_V$, which is 27.2.2.1 (in the middle equality we used the definition of $\varphi_{U,V}$, plus $(\theta')^V_V=\text{id}_{\mathcal{F}'_V}$)."

Potential typo in the statement of the lemma: it is written "[l]et $U\subset X$ be an open", but condition (3) is that $U\cap p(i(Z))=\emptyset$, which only makes sense if $U\subset S$.

Minor style typo: in the statement perhaps one would want to enclose the name of the categories between curly braces to fit the style in 29.11.6.

2nd proof (maybe worth of mention?): The functor $\underline{\text{Spec}}_S$ is fully faithful by remark (ii) of https://stacks.math.columbia.edu/tag/01LQ#comment-8438 , and it is essentially surjective by 29.11.3.

Final remarks on why I am writing all these comments in this section instead of just introducing a pull request on GitHub: In my opinion, it would be better to change this whole section to the study of the representability of $G$ instead of $F$, and maybe mentioning the existence of $F$ just at the end plus the categorical remark in #8435. I do think so because (a) $G$ is easier to remember and more motivated than $F$, see #8435 (I also think it's what #3430 had on his mind), and (b) proofs with $G$ are less wordy than those with $F$. However, I was unsure whether such a big edit would be accepted or if only a proper subset of all the edits would pass. In consequence, I deemed it better to point it all out here.

I think it would be nice to add the following two remarks somewhere in this section. I will be adopting the POV explained in #8435 of representability of $G$ instead of $F$.

(i) So far, we have an explicit description of the map $$\tag{1}\label{map} \text{Hom}_S(T,\underline{\text{Spec}}_S\mathcal{A}) \to\operatorname{Hom}_{\mathcal{O}_S\text{-Alg}}(\mathcal{A},f_*\mathcal{O}_T),$$ Namely, it is the one induced by the universal element $\xi=(\mathcal{A}\to\pi_*\mathcal{O}_{\underline{\text{Spec}}_S\mathcal{A}})\in G(\pi)$. However, we don't have yet an explicit description of the inverse of \eqref{map}. Here's one: Given an $\mathcal{O}_S$-linear map $\mathcal{A}\to f_*\mathcal{O}_T$, the induced map $T\to\underline{\text{Spec}}_S\mathcal{A}$ is the unique morphism over $S$ such that for all open affine $U\subset S$, the restricted map over $U$ equals $f^{-1}(U)\to\operatorname{Spec}(\mathcal{A}(U)) \xrightarrow{i_U^{-1}}\pi^{-1}(U)$, where $i_U$ is the map in 27.3.4 and the first arrow comes from Schemes, 26.6.4. Proof: First we note that $U\subset S\mapsto\text{Hom}_U(f^{-1}(U),\pi^{-1}(U))$ is a sheaf on $S$ (basically because $U\subset S\mapsto\text{Hom}_U(f^{-1}(U),\underline{\text{Spec}}_S\mathcal{A})$ is a sheaf). Then we note that for any open $U\subset S$, we have an induced map $$\text{Hom}_U(f^{-1}(U),\pi^{-1}(U))\to\text{Hom}_{\mathcal{O}_U\text{-Alg}}(\mathcal{A}|_U,\underbrace{f_*\mathcal{O}_T|_U}_{(f|_{f^{-1}(U)})_*\mathcal{O}_{f^{-1}(U)}}).$$ Namely, it is the one obtained by the universal element $\iota^*\xi=(\mathcal{A}|_U\to(\pi_*\mathcal{O}_{\underline{\text{Spec}}_S\mathcal{A}})|_U)$ of $G_\iota$, where $\iota:\pi^{-1}(U)\to\underline{\text{Spec}}_S\mathcal{A}$ is the inclusion and we are using the notation given in https://stacks.math.columbia.edu/tag/001L#comment-8427 . From here, it is not difficult to verify that the diagram $$\require{AMScd} \begin{CD} \text{Hom}_S(T,\underline{\text{Spec}}_S\mathcal{A})@>>> \operatorname{Hom}_{\mathcal{O}_S\text{-Alg}}(\mathcal{A},f_*\mathcal{O}_T)\\ @VVV@VVV\\ \text{Hom}_U(f^{-1}(U),\pi^{-1}(U))@>>> \text{Hom}_{\mathcal{O}_U\text{-Alg}}(\mathcal{A}|_U,f_*\mathcal{O}_T|_U) \end{CD}$$ commutes. In other words, we have a morphism of sheaves $\text{Hom}_{(-)}(f^{-1}(-),\pi^{-1}(-))\to\mathcal{H}om_{\mathcal{O}_S\text{-Alg}}(\mathcal{A},f_*\mathcal{O}_T)$ on $S$. Thus, we can characterize the inverse of \eqref{map} locally on $S$ and we finish by the proof in https://stacks.math.columbia.edu/tag/01LT#comment-8436

(ii) An explanation of why $\underline{\text{Spec}}_S$ is a functor: If we have a morphism of $\mathcal{O}_S$-algebras $\mathcal{A}\to\mathcal{B}\cong\pi_{\mathcal{B},*}\mathcal{O}_{\underline{\text{Spec}}_S\mathcal{B}}$ (we've used 27.4.6, (3)) then we obtain a canonical map $\underline{\text{Spec}}_S\mathcal{B}\to\underline{\text{Spec}}_S\mathcal{A}$ over $S$. By remark (i), it is the unique morphism of schemes over $S$ such that for every open affine $U\subset S$, we have that $\pi_\mathcal{B}^{-1}(U)\to\pi_\mathcal{A}^{-1}(U)$ identifies with $\text{Spec}(\mathcal{B}(U))\to\text{Spec}(\mathcal{A}(U))$ via the isomorphisms of Lemma 27.3.4. It follows that $\underline{\text{Spec}}_S$ is a contravariant functor from the category of quasi-coherent $\mathcal{O}_S$-algebras to the category of schemes over $S$. On the other hand, equation \eqref{map} specializes to $$\text{Hom}_S(\underline{\text{Spec}}_S\mathcal{B}, \underline{\text{Spec}}_S\mathcal{A}) \xrightarrow{\cong}\operatorname{Hom}_{\mathcal{O}_S\text{-Alg}}(\mathcal{A},\mathcal{B}),$$ so the functor is fully faithful.

'The second main result is that for a morphism $f$ if ringed sites....' right above the displayed equality of derived completions has a typo. The 'if' should be 'of'.

I would propose writing the whole proof in this way: I'm going to adopt the POV of representability of $G$ instead of $F$, see https://stacks.math.columbia.edu/tag/01LQ#comment-8435 . We have $\mathcal{A}=\widetilde{A_R}$ (where $A_R$ just means $A$ regarded as an $R$-module, notation introduced in the paragraph before 10.14.3). To show that $\pi:\operatorname{Spec} A\to\operatorname{Spec} R$ represents $$\begin{align*} G:\text{Sch}^\text{opp}/S&\to\text{Sets}\\ (f:T\to S)&\mapsto \operatorname{Hom}_{\mathcal{O}_S\text{-Alg}}(\mathcal{A},f_*\mathcal{O}_T), \end{align*}$$ we check that the identity map $\varphi_\text{univ}:\widetilde{A_R}\to\pi_*\widetilde{A}=\widetilde{A_R}$ is a universal element for $G$. Fix $f:T\to S$. We have a map $\text{Hom}_S(T,\operatorname{Spec}A)\to G(f)$, induced by $\varphi_\text{univ}$. To construct the inverse, for an $\mathcal{O}_S$-linear map $\varphi:\widetilde{A_R}\to f_*\mathcal{O}_T$ (an element of $G(f)$), we have a map on global sections $A\to\Gamma(T,\mathcal{O}_T)$, which is $R$-linear. Therefore, it induces a morphism $T\to\operatorname{Spec}A$ over $\operatorname{Spec} R$, by Schemes, 26.6.4. On the one hand, it is easy to see that the composite $\text{Hom}_S(T,\operatorname{Spec}A)\to G(f)\to\text{Hom}_S(T,\operatorname{Spec}A)$ is the identity (use uniqueness in Schemes, 26.6.4). On the other hand, to see that $G(f)\to \text{Hom}_S(T,\operatorname{Spec}A)\to G(f)$ is the identity, we note that all the involved maps are compatible with restrictions to distinguished open subsets of $S$ (i.e., given $U\subset S$ distinguished open, one restricts to $f^{-1}(U)\to U$ and to $\pi^{-1}(U)\to U$). Hence, it suffices to see that $$\begin{matrix} G(f)&\longrightarrow&\text{Hom}_S(T,\operatorname{Spec}A)&\longrightarrow& G(f)\\ &\searrow&&\swarrow\\ &&\text{Hom}_{R\text{-Alg}}(A,\Gamma(T,\mathcal{O}_T)) \end{matrix}$$ commutes. But this is easy.